Question

If a,b,c are in continued proportion prove that :
i) $\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a}{c}$.
ii) $\frac{a^4+a^2{b}^2+b^4}{b^4+b^2{c}^2+c^4}=\frac{a^2}{c^2}$.

Answer 1

Here $\frac{a}{b}=\frac{b}{c}\therefore b^2=ac$

$i)LHS=\frac{a^2+ab+b^2}{b^2+bc+c^2}=\frac{a^2+ab+ac}{ac+bc+c^2}$

[Putting $b^2=ac$]

$=\frac{a(a+b+c)}{c(a+b+c)}=\frac{a}{c}=RHS$

$ii)LHS=\frac{a^4+a^2{b}^2+b^4}{b^4+b^2{c}^2+c^4}=\frac{a^4+a^2{b}^2+(ac{)}^2}{(ac{)}^2+b^2{c}^2+c^4}=\frac{a^4+a^2{b}^2+a^2{c}^2}{a^2{c}^2+b^2{c}^2+c^4}=\frac{a^2(a^2+b^2+c^2)}{c^2(a^2+b^2+c^2)}=\frac{a^2}{c^2}=RHS$